# Strict Ordering on Integers is Asymmetric

## Theorem

Let $\eqclass {a, b} {}$ denote an integer, as defined by the formal definition of integers.

Then:

 $\ds \eqclass {a, b} {}$ $<$ $\ds \eqclass {c, d} {}$ $\, \ds \implies \,$ $\ds \eqclass {c, d} {}$ $\nless$ $\ds \eqclass {a, b} {}$

That is, strict ordering on the integers is asymmetric.

## Proof

By the formal definition of integers, we have that $a, b, c, d, e, f$ are all natural numbers.

To eliminate confusion between integer ordering and the ordering on the natural numbers, let $a \prec b$ denote that the natural number $a$ is less than the natural number $b$.

We have:

 $\ds \eqclass {a, b} {}$ $<$ $\ds \eqclass {c, d} {}$ $\ds \leadsto \ \$ $\ds a + d$ $\prec$ $\ds b + c$ Definition of Strict Ordering on Integers $\ds \leadsto \ \$ $\ds b + c$ $\nprec$ $\ds a + d$ Definition of Ordering on Natural Numbers $\ds \leadsto \ \$ $\ds \eqclass {c, d} {}$ $\nless$ $\ds \eqclass {a, b} {}$ Definition of Strict Ordering on Integers

$\blacksquare$